QWidget::isVisible() not working on linux

  • when I test fWidget->isVisible() in centos , It always returns false even if the fWidget is shown . While I use !fWidget->isHidden(); it works . What's the problem ? anyone knows the reason and meets the same problem before ? thanks ~~

  • Is this a top-level widget?

  • no , it is not top-level widget . But I am sure it is shown . Actually it is a QGLWidget() , and I draw a lot of models inside it .

  • Can you give us a small example which reproduce the problem?

    inline bool QWidget::isVisible() const
    { return testAttribute(Qt::WA_WState_Visible); }

    inline bool QWidget::isHidden() const
    { return testAttribute(Qt::WA_WState_Hidden); }

  • thank you , guys , but I can't give a example , because this is what I met in our current project , the
    project is very large . I wonder whether this is because the fWidget is not a window , it is the child of some widget , I tested using isVisble on the child of visible widgets , it can work . so It is really strange why it can not work here .

  • The infomation you provited is too limited, so I think other people cannot give you some useful hint.

    If you are unsure whether it related to something,you can write a small program to test it.

  • thank you , MR 2 , I have tested it with the small program , but didn't find any strange result . So thank you again , Mr 2 , just ignore this .

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